Question

The actual time it takes to cook a ten pound turkey is a normally distributed. Suppose that a random sample of 19 ten pound turkeys is taken. Given that an average of 2.9 hours and a standard deviation of .24 hours was found for a sample of 19 turkeys, calculate a 95% confidence interval for the average cooking time of a ten pound turkey.

Answer 1

Solution :

Given that,

= 2.9

s = 0.24

n = 19

Degrees of freedom = df = n - 1 = 19 - 1 = 18

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,18 =2.101

Margin of error = E = t_/2,df * (s /n)

= 2.101 * (0.24 / 19)

= 0.11

Margin of error = 0.11

The 95% confidence interval estimate of the population mean is,

- E <  < + E

2.9 - 0.11 < < 2.9 + 0.11

2.78 < < 3.01

(2.78, 3.01 )